INEQUALITY'S BLOG

August 1, 2010

Problem 038

Filed under: Uncategorized — KKKVVV @ 6:44 am

Let f:R\longrightarrow R_{0}^{+} and f''>0. For all 0<a\leq b\leq c ; a, b, c\in I ; Prove that:

a^2f(b)+b^2f(c)+c^2f(a)\geq abf(c)+bcf(a)+caf(b)

Proof

Let m=c, n=a ; p=b but we have a\leq b\leq c\Longrightarrow m\geq n; p\geq n

From the inequality by van khea for m\geq n; p\geq n ; a\leq b\leq c and f''>0 we have

m(c-b)f(a)-n(c-a)f(b)+p(b-a)f(c)\geq 0

\Longrightarrow c(c-b)f(a)-a(c-a)f(b)+b(b-a)f(c)\geq 0

\Longrightarrow a^2f(b)+b^2f(c)+c^2f(a)\geq abf(c)+bcf(a)+caf(b)

Therefor the proof is completed.

 

Van Khea    

 

See also the inequality by Van Khea

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